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Analytical Geometry
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sowjanya gudipati
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Joined: 13 Oct 2007
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14 Jan 2008 12:59:56 IST
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Re:circles
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14 Jan 2008 13:17:18 IST
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since eqn of circle is x^2+y^2-2rx-2hy+h^2=0
comparing it by standard eqn ,x^2+y^2+2gx+2fy+c=0
we get
it's centre as (r,h) and radius as r
now ,let the eqn of tangent be y=mx+c.
since it is passing through origin ,eqn will reduce to y=mx.
now
length of perpendicular from centre of circle to tangent = radius of circle.
(mr-h)/
1+m^2 =r.
1+m^2 =r.therefore from this u obtain m=(h^2-r^2)/2rh
eqn of tangent will be y={(h^2-r^2)/2rh}x
rate me if correct











